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The length of a rectangular poster is 9 more inches than half its width.The area of the poster is 140 square inches

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Answer:

width = 10 inches; length = 14 inches

Explanation:

  • It appears that you're trying to find the dimensions of the poster given the info about the length and the width.
  • If this is the case, my work below will help you find these dimensions.
  • If you're trying to find something else, write it in the comments and I'll edit my answer to help you.

The formula for the area of a rectangle:

The formula for the area of a rectangle is given by:

A = lw, where

  • A is the area in square units,
  • l is the length,
  • and w is the width

Relating length and width:

Since we're told that the poster's length is 9 more inches than half its width, we can model this with the following equation:

l = 1/2w + 9

Finding the width:

Now we can substitute 140 for A and 1/2w + 9 for w to find w, the width of the poster:

140 = (1/2w + 9)w

140 = 1/2w^2 + 9w

Putting the 140 = 1/2w^2 + 9w in standard form (ax^2 + bx + c = 0)

We can subtract 1/2w^2 and 9w from both sides to put the quadratic in standard form, which will allow us to solve it:

(140 = 1/2w^2 + 9w) - 1/2w^2 - 9w

-1/2w^2 - 9w + 140 = 0

Clear the fraction in -1/2w^2 to simplify the quadratic:

In order to clear the -1/2 from -1/2w^2, we can divide both sides both sides by -1/2 (note that when dividing by a fraction, we multiply by the reciprocal, so we'll end up both sides multiplying by -2:

(-1/2w^2 - 9w + 140 = 0) / -1/2

(-1/2w^2 - 9w + 140 = 0) * -2

1w^2 + 18w - 280 = 0

w^2 + 18w - 280 = 0

Thus, our a value is 1, our b value is 18, and our c value is -280.

Solving w^2 + 18w - 280 = 0 using the quadratic formula:

Now we can solve the quadratic using the quadratic formula, which is given by:

x = -b / 2a ±( √(b^2 - 4ac)) / 2a, where

  • x is a solution to the quadratic,

Since we're solving for w, we can swap x with w in the quadratic equation to prevent confusion:

Thus, we can plug in 1 for a, 18 for b, and -280

w = -18 / 2(1) ± (√((18)^2 - 4(1)(-280))) / 2(1)

w = -18 / 2 ± (√1444) / 2

w = -9 ± 38/2

w = -9 + 19 and x = -9 - 19

w = 10 and x = -28

Thus, w = 10 and w = -28.

Since we can't have a negative dimension, the width is 10 inches.

Determine the length:

Now we can find the poster's length by plugging in 140 for A and 10 for w in the rectangle area formula and solving for l:

(140 = 10l) / 10

14 = l

Thus, the poster's length is 14 inches.

Check the validity of the answer:

We can check that we've correctly determined the poster's dimensions by seeing if the product of 10 and 14 is 140 and if half of 10 plus 9 = 14.

10 * 14 = 140

1/2 * 10 = 5

5 + 9 = 14

Thus, our answers are correct.

User Josh Morrison
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